2963. 宝石机关
摘要
Title: 2963. 宝石机关
Tag: 模拟
Memory Limit: 64 MB
Time Limit: 1000 ms
Powered by:NEFU AB-IN
2963. 宝石机关
-
题意
略
-
思路
思路就是开三个哈希表，先让第一个哈希表把p的所有数打上表示，当p[i]遍历到时，判断a - p[i]和 b - p[i]是否存在，如果存在就抹去第一个的标记，去标记A或B数组
-
代码
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73'''
Author: NEFU AB-IN
Date: 2023-06-09 23:46:37
FilePath: \LanQiao\2963\2963.py
LastEditTime: 2023-06-10 00:31:47
'''
# import
from sys import setrecursionlimit, stdin, stdout, exit
from collections import Counter, deque
from heapq import heapify, heappop, heappush, nlargest, nsmallest
from bisect import bisect_left, bisect_right
from datetime import datetime, timedelta
from string import ascii_lowercase, ascii_uppercase
from math import log, gcd, sqrt, fabs, ceil, floor
class sa:
def __init__(self, x, y):
self.x = x
self.y = y
def __lt__(self, a):
return self.x < a.x
# Final
N = int(1e5 + 10)
M = 20
INF = int(2e9)
# Define
setrecursionlimit(INF)
input = lambda: stdin.readline().rstrip("\r\n") # Remove when Mutiple data
read = lambda: map(int, input().split())
LTN = lambda x: ord(x.upper()) - 65 # A -> 0
NTL = lambda x: ascii_uppercase[x] # 0 -> A
# —————————————————————Division line ——————————————————————
p = [0] * N
n, a, b = read()
p[1:] = read()
A, B, d = Counter(), Counter(), Counter()
for i in range(1, n + 1):
d[p[i]] = 1
for i in range(1, n + 1):
if d[p[i]] == 0:
continue
if d[a - p[i]] == 1 or A[a - p[i]] == 1:
A[a - p[i]] = 1
A[p[i]] = 1
d[a - p[i]] = 0
d[p[i]] = 0
elif d[b - p[i]] == 1 or B[b - p[i]] == 1:
B[b - p[i]] = 1
B[p[i]] = 1
d[b - p[i]] = 0
d[p[i]] = 0
else:
print("NO")
exit(0)
print("YES")
for i in range(1, n + 1):
if A[p[i]]:
print(0, end = " ")
else:
print(1, end = " ")